The article Cryptographic Protocols with Everyday Objects by James Heather, Steve Schneider, and Vanessa Teague describes the following dating protocol (due to Bert den Boer):

Alice and Bob wish to determine whether they both want to go on a date; but they want to avoid the embarrassing situation in which one of them does not want to go on a date, but knows that the other would have liked to do so. Essentially they need a two-player veto protocol: they want to compute whether at least one has vetoed the date, without revealing any further information.

Q: Bennett’s solution uses playing cards. Does this problem admits cryptographic solution?

Of course it can be reduced to Yao's Millionaires' Problem. But probably this problem has a simpler solution.

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    $\begingroup$ What you need is a secure two party computation protocol to compute an AND. $\endgroup$ – SEJPM Feb 9 '18 at 14:13
  • $\begingroup$ You can reduce it to socialist millionaires, which is pretty simple. $\endgroup$ – CodesInChaos Feb 9 '18 at 16:19
  • $\begingroup$ @CodesInChaos Does the socialist millionaire problem account for the age of all parties? $\endgroup$ – Q-Club Feb 9 '18 at 19:01
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    $\begingroup$ @Q-Club I strongly condemn age discrimination. $\endgroup$ – CodesInChaos Feb 9 '18 at 19:16
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    $\begingroup$ I'm surprised to see the dating protocol attributed to Charles Bennett, because it is known as Bert den Boer's five card trick. Den Boer used it as a very nice and playful introduction to his Eurocrypt '89 about MPC protocols for match-making. However, Bennett is acknowledged by Claude Crépeau and Joe Kilian in their Crypto '93 paper "Discreet Solitary Games," which is about card-based MPC protocols for the Secret Santa problem. So Bennett was a bit into these protocols as well, and probably liked Den Boer's five-card trick a lot. $\endgroup$ – Berry Schoenmakers Feb 24 '19 at 22:58

You can use a protocol solving the Socialist millionaires problem for this. Socialist millionaires compares two integers for equality. There are relatively simple implementations of this protocol, similar to Diffie-Hellman, which are used to implement PAKE (password authenticated key-exchange).

Agree a fixed integer to denote true. A party chooses that value if they want to signal true and a random other value (say 256-bits) if they want to signal false. This clearly works for the case where at least one party chooses true. If both parties choose false it will almost certainly produce a not-equal result from SM, since the chance of two random values being equal is negligible.

Thus you can securely implement the AND function using a Socialist millionaires protocol.

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First agree on a large prime $p$, a generator $g$, and a random value module $p$ to represent $true$. Each party has their choice of date or not as $C_{a,b} \in \{0,1\}$. Then each party $i$ computes:

$ \\ false \leftarrow random() \\ k_{private} \leftarrow random() \\ choice_i \leftarrow H(true \times C_i + false \times (1 - C_i)) \\ e \leftarrow g^{choice_i \times k_{private}} \\ \text{Send } e_i \text{ to the other player, receiving } e_{\bar{i}} \\ \text{Publish } e_{\bar{i}}^{k_{private}} $

This is just a typical private set intersection specialized to a single element.

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  • $\begingroup$ This doesn't seem secure. Even if I vote false. I still know if you voted true or not. $\endgroup$ – Meir Maor Feb 10 '18 at 5:46
  • $\begingroup$ Claiming you know and showing the math are two different things. Care to show how? This is a typical protocol. I sent $g^{ck_a}$ and you don't have $k$. You can get me to compute $g^{ck_ak_b}$ but then your $c$ must be 'true' and revealing the result of an and operation is the whole point. $\endgroup$ – Thomas M. DuBuisson Feb 10 '18 at 6:09
  • $\begingroup$ I take it back, I was confusing choice for Ci $\endgroup$ – Meir Maor Feb 10 '18 at 6:46
  • $\begingroup$ What is H function here? $\endgroup$ – Rogach Feb 14 '18 at 6:22
  • $\begingroup$ Just a hash function. SHA512 if you'd like. $\endgroup$ – Thomas M. DuBuisson Feb 14 '18 at 6:45

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